Diving hawks.

3. When the Sun is directly overhead, a hawk dives toward the ground with a constant velocity of 5.00 m/s at 60.0 degrees below the horizontal. Calculate the speed of her shadow on the level ground.

I am having some trouble picturing this problem in terms of a coordinate grid (this is in a chapter dealing with motion in two dimensions). I'm basically making a triangle, and I know (or at least I'm hoping) that the shadow is the horizontal component of the vector that is the hawk's flight. If it's 60 degrees below the "horizontal" or, the way I see it, the "line of sight", the direction is equivalent to 300 degrees if making an imaginary coordinate grid at the point before she descends, making the h.c. = 5.00cos300. Or, actually, the entire thing is a 30-60-90 triangle, so the velocity of her shadow is 2.5 m/s.

You should get the same answer if you did the problem with the cosine of 300 degrees or with the 30-60-90 triangle. Both descriptions of the situation can accurately be used to find the answer.

Also, your thinking, as far as I understand it, is correct. 60 degrees below the horizontal in the can be described by an angle of 300 degrees. Was this all you were asking or is there more to your train of thought that you were unsure about?